Question

How do you solve `2x^2+5x-3<=0`?

Answer 1

`-3<=x<=1/2`

Explanation:

First find the roots of the answer
`2x^2+5x-3`=`(2x-1)(x+3)`
graph{(2x-1)(x+3) [-5.546, 5.55, -2.773, 2.774]}
Since the equation is below the x-axis,
the answer will be between `-3 and 1/2`
Therefore `-3<=x<=1/2`

Answer 2

`-3<=x<=1/2`

Explanation:

`"factorise the quadratic"`

`(x+3)(2x-1)<=0`

`"solve "(x+3)(2x-1)=0`

`rArrx=-3" or "x=1/2larrcolor(blue)"zeros"`

`"the coefficient of "x^2" term ">0rArruuu`

`rArr-3<=x<=1/2`
graph{2x^2+5x-3 [-10, 10, -5, 5]}